Let $\psi : G\times M\rightarrow M$ be a smooth left action of a lie group $M$. For $A\in Lie(G)$, we define $X_A\in \mathfrak{X}(M)$ be the fundamental vector field $X_A(p):=\frac{d}{dt}exp(tA).p|_{t=0}$. Now my goal is to show that $[X_A,X_B]=-X_{[A,B]}$ , so that we have an anti-homomorphism of lie algebras, here's what I have done :
$[X_A,X_B](p).f=\frac{d}{dt}(f(\phi_{X_B}^{-\sqrt{t}}\circ \phi_{X_A}^{-\sqrt{t}}\circ \phi_{X_B}^{\sqrt{t}} \circ \phi_{X_A}^{\sqrt{t}}(p)))|_{t=0}$
$=\frac{d}{dt}(f(\exp(-\sqrt{t}B)\exp(-\sqrt{t}A)\exp(\sqrt{t}B)\exp(\sqrt{t}A).p))|_{t=0}$
And
$-X_{[A,B]}(p).f=\frac{d}{dt}f(\exp(-t[A,B]).p)|_{t=0}=\frac{d}{dt}f(\exp(t[B,A]).p)|_{t=0}$ $=\frac{d}{dt}(f(\phi_A^{-\sqrt{t}})\circ \phi_B^{-\sqrt{t}}\circ \phi_A^{\sqrt{t}} \circ \exp(\sqrt{t}B).p))|_{t=0}$ $=\frac{d}{dt}(f(\exp(\sqrt{t}B)\exp(\sqrt{t}A)\exp(-\sqrt{t}B)\exp(-\sqrt{t}A).p))|_{t=0}$
Where I have used the fact $\phi_B^t(p)=p\exp(tB)$. And I used the fact, hope it's true , that $exp(t[B,A])=\phi_A^{-\sqrt{t}}\circ \phi_B^{-\sqrt{t}}\circ \phi_A^{\sqrt{t}} \circ \phi_B^{\sqrt{t}}(e)$
so I must be making some mistake since the signs don't quite add up but I can't quite figure out what it is.
It's also weird because if we consider the right action we will have that $[X_A,X_B]=X_{[A,B]}$, and this using the same methods of using the flows I was able to prove like this :
$\phi_{X_B}^{-\sqrt{t}}\circ \phi_{X_A}^{-\sqrt{t}}\circ \phi_{X_B}^{\sqrt{t}} \circ \phi_{X_A}^{\sqrt{t}}(p)=p.\exp(\sqrt{t}A)\exp(\sqrt{t}B)\exp(-\sqrt{t}A)\exp(-\sqrt{t}B)$
$\exp(t[A,B])=\phi_B^{-\sqrt{t}}\circ \phi_A^{-\sqrt{t}}\circ \phi_B^{\sqrt{t}}\circ \phi_A^{\sqrt{t}}(e)$
$=\exp(\sqrt{t}A)\exp(\sqrt{t}B)\exp(-\sqrt{t}A)\exp(-\sqrt{t}B)$
So I am confused where I could have made a mistake in the previous case but possibly not on this one.
Any help is aprecciated. Thanks in advance.